oxygen is acting as an oxidizing agent in all of the following reactions except\n\na. $2 \\text{c(s)} +…
oxygen is acting as an oxidizing agent in all of the following reactions except\n\na. $2 \\text{c(s)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{co(g)}$\n\nb. $\\text{s(s)} + \\text{o}_2\\text{(g)} \\rightarrow \\text{so}_2\\text{(g)}$\n\nc. $2 \\text{f}_2\\text{(g)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{of}_2\\text{(g)}$\n\nd. $2 \\text{na(s)} + \\text{o}_2\\text{(g)} \\rightarrow \\text{na}_2\\text{o}_2\\text{(s)}$\n\ne. $2 \\text{mg(s)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{mgo(s)}$
Answer
Explanation:
Step1: Define oxidizing agent
An oxidizing agent gains electrons and undergoes reduction, meaning its oxidation state decreases.
Step2: Analyze reaction A
In $2\text{C} + \text{O}_2 \rightarrow 2\text{CO}$, oxygen goes from $0$ to $-2$. It is an oxidizing agent.
Step3: Analyze reaction B
In $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$, oxygen goes from $0$ to $-2$. It is an oxidizing agent.
Step4: Analyze reaction C
In $2\text{F}_2 + \text{O}_2 \rightarrow 2\text{OF}_2$, fluorine is more electronegative than oxygen. Fluorine goes from $0$ to $-1$, and oxygen goes from $0$ to $+2$. Oxygen is oxidized, making it a reducing agent.
Step5: Analyze reaction D
In $2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2$, oxygen goes from $0$ to $-1$. It is an oxidizing agent.
Step6: Analyze reaction E
In $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$, oxygen goes from $0$ to $-2$. It is an oxidizing agent.
Answer:
C. $2\text{F}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{OF}_2(\text{g})$